0
$\begingroup$

Say you have a charged particle in a region that contains a fluid that will produce a drag force that goes as $F=-kv$ where $v$ is the speed and $k$ is some constant. The region also contains a uniform magnetic field. Suppose you give the particle some initial velocity $v_0$ in the plane perpendicular to the magnetic field. What will be the particle's subsequent motion? Please provide semi-quantitative answers. Note that this is not a homework question.

$\endgroup$
2
  • 1
    $\begingroup$ What have you attempted so far? $\endgroup$ Dec 25 '13 at 0:00
  • $\begingroup$ My guess is that it is a logarithmic spiral based on the fact that the speed falls off exponentially (from the drag force), although I'm not sure how to show it analytically. $\endgroup$
    – zeta
    Dec 25 '13 at 0:05
4
$\begingroup$

Let's find the complete solution of the problem.

A complete solution of the problem would be the solution to the linear ODE,

$m \dot{\mathbf{v}} =q\mathbf{v} \times \mathbf{B}-k \mathbf{v}$

Assume without loss of generality that the magnetic field is pointed along the z-axis,so $\mathbf{B} = B \mathbf{\hat{z}}$. So our equation simplifies to,

$m \dot{\mathbf{v}} =qB\mathbf{v} \times \mathbf{\hat{z}}-k \mathbf{v}$

Dividing both sides of the equation by $m$ and for simplicity in the notation, let $\omega=\frac{qB}{m}$ and $\gamma=\frac{k}{m}$.

So,$\dot{\mathbf{v}} =\omega\mathbf{v} \times \mathbf{\hat{z}}-\gamma \mathbf{v}$

Using $\mathbf{v}=\begin{pmatrix} v_{x}\\v_{y}\\v_{z} \end{pmatrix}$ and writng the given eqaution in matrix form we have,

$\dot{\mathbf{v}} =\begin{pmatrix} -\gamma & \omega & 0 \\ -\omega & -\gamma & 0 \\ 0 & 0 & -\gamma \end{pmatrix} \mathbf{v}=A \mathbf{v}$

This is linear ODE which can be solved using the matrix exponenetial as,

$\mathbf{v}=e^{At} \mathbf{v_0}$ To simply this equation we can find the eigenvaules of A,use a similarity transform to convert it to a diagonal matrix which this greatly simplifies the matrix exponential.

The eigenvalues and the corresponding eigenvectors are,

$\lambda_{1}=-\gamma ,\mathbf{v_1}=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

$\lambda_{2}=-\gamma-i\omega ,\mathbf{v_2}=\begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}$

$\lambda_{3}=-\gamma+i\omega ,\mathbf{v_3}=\begin{pmatrix} i \\ 1 \\ 0 \end{pmatrix}$

Using $S=[ \mathbf{v_1} \mathbf{v_2} \mathbf{v_3}]$ and performing a similarity transform on the matrix A,$S^{-1}AS=D$ where D is diagonal matrix with the eigenvalues as the diagonal elements.And here we witness the power of similarity transformations as,

$e^{At}= S \begin{pmatrix} e^{\lambda_{1}t} &0 &0 \\0 & e^{\lambda_{2}t} &0 \\ 0& 0& e^{\lambda_{3}t} \end{pmatrix} S^{-1}$

(After some tedious calculations ans using $e^{ix}=\cos{x}+isin{x}$)

$=\begin{pmatrix} e^{-\gamma t}\cos(\omega t) &e^{-\gamma t}\sin(\omega t) &0\\ -e^{-\gamma t}\sin(\omega t) &e^{-\gamma t}\sin(\omega t) &0\\0 & 0 &e^{-\gamma t}\end{pmatrix}$

Therefore, $\mathbf{v}=\begin{pmatrix} e^{-\gamma t}\cos(\omega t) &e^{-\gamma t}\sin(\omega t) &0\\ -e^{-\gamma t}\sin(\omega t) &e^{-\gamma t}\sin(\omega t) &0\\0 & 0 &e^{-\gamma t}\end{pmatrix} \mathbf{v_0}$

Writing out the components,

$v_{x}=e^{-\gamma t}(v_{x_0} \cos{\omega t}+v_{y_0} \sin{\omega t})$

$v_{x}=e^{-\gamma t}(-v_{x_0} \sin{\omega t}+v_{y_0} \cos{\omega t})$

$v_{x}=e^{-\gamma t} v_{z_0}$

This the just the equation of a helix with both the pitch and radius decreasing exponentially with $\gamma$.However,the angular frequency is the same as that without drag,$\omega$.

Here is a sample trajectory, enter image description here

$\endgroup$
0
$\begingroup$

The magnetic field does no work, so it does not change the speed of the particle. The drag force results in an exponential decrease in speed. If the drag force is very small, one would expect the particle to move in a circle with radius $r = \frac{mv}{qB}$, where $v = v_0 e^{-\frac{k}{m} t}$, i.e., a spiral.

A precise solution would proceed from

$$m\dot{\mathbf{v}} = q B \, \mathbf{v} \times \hat{\mathbf{z}} - k \,\mathbf{v}$$

This is a set of first-order, linear, coupled ODE's. Not very difficult to solve. In particular, $$\dot{\mathbf{v}} = M \,\mathbf{v}$$ where $M$ is a matrix, so you would start by finding its eigenvectors.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.